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Question
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Solution
\[Let\ I = \int_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} d x ................(1)\]
\[ = \int_a^b \frac{f\left( a + b - x \right)}{f\left( a + b - x \right) + f\left( a + b - a - b + x \right)} d x\]
\[ = \int_a^b \frac{f\left( a + b - x \right)}{f\left( a + b - x \right) + f\left( x \right)} d x\]
\[ \therefore I = \int_a^b \frac{f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} d x ...............(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_a^b \left[ \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} + \frac{f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} \right] d x\]
\[ = \int_a^b \frac{f\left( x \right) + f\left( a + b - x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx\]
\[ = \left[ x \right]_a^b \]
\[ = b - a\]
\[\text{Hence, }I = \frac{b - a}{2}\]
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